Note that this algorithm consumes constant heap space, and has an n factorial time complexity, where n is the length of the input list
Print the results on the run:
global result
result = []
def permutation(li):
if li == [] or li == None:
return
if len(li) == 1:
result.append(li[0])
print result
result.pop()
return
for i in range(0,len(li)):
result.append(li[i])
permutation(li[:i] + li[i+1:])
result.pop()
Example:
permutation([1,2,3])
Output:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]